We define \(F^{n+1}\)  iteratively by the following scheme:
\(\begin{split}\partial _tF^{n+1}+v\cdot \nabla _xF^{n+1} &= \frac{1}{\tau (F^n)}(\mathcal {S}_{Pr}(F^n)-F^{n+1}), \cr F^{n+1}(x,v,0) &= F_0(x,v),\end{split}\) 

with \(F^0(x,v,t)=F_0(x,v)\) . We use induction argument. We assume the statement (1) - (4) for \(n\) -th step.
We first observe that
\(\mathcal {S}_{Pr}(F^n) &= \frac{\rho _n}{\sqrt{2\pi T_n}^3} \left[ 1+\frac{1-Pr}{5}\frac{q_n\cdot v}{\rho _n T_n^2} \left(\frac{|v|^2}{2T_n}-\frac{5}{2}\right)\exp \left(-\frac{|v|^2}{2T_n}\right)\right]\cr &\ge \mathcal {M}(F^n)\left(1-C\frac{1-Pr}{5}\frac{|q_{n}|}{\rho _{n} T_{n}^{3/2}}\right),\) 

where we used
\(\bigg |x^m\exp \left(-\frac{x^2}{2T}\right)\bigg | \le C T^{m/2}.\) 

Applying Lemma REF  and the induction hypothesis yields
\(\mathcal {S}_{Pr}(F^n)\ge \mathcal {M}(F^n)\left(1-C\sqrt{\mathcal {E}(f^n)}\right)\ge \mathcal {M}(F^n)\left(1-CM_0\right).\) 

Therefore for sufficiently small \(M_0\) , we have \(\mathcal {S}_{Pr}(F^n)\ge 0\) . Then the non-negativity of \(F^n\)  follows directly from the mild formulation of (REF ):
\(F^{n+1}(x,v,t) = e^{-\int _0^t\frac{1}{\tau (F^n)}dt}F_0(x-vt,v)\cr +\frac{1}{\tau (F^n)}e^{-\int _s^t\frac{1}{\tau (F^n)}dt}\int _0^{T_*}\mathcal {S}_{Pr}(F^n)(x+v(s-t),v,s)ds.\) 
Now we prove the uniform boundedness of the energy norm. For this, we substitute \(F^{n+1}=m+\sqrt{m}f^{n+1}\)  into (REF ) to get
\(\partial _tf^{n+1}+v\cdot \nabla _xf^{n+1} +\frac{1}{\tau _0}f^{n+1} &= \frac{1}{\tau _0}P_{Pr}(f^n)+\Gamma (f^n), \cr f^{n+1}(x,v,0) &= f_0(x,v),\) 

where \(f_0^n(x,v)=(F_0(x,v)-m)/\sqrt{m}\) . Taking \(\partial ^{\alpha }_{\beta }\)  on both sides:
\(\partial _t\partial ^{\alpha }_{\beta } f^{n+1}+v\cdot \nabla _x \partial ^{\alpha }_{\beta } f^{n+1} +\frac{1}{\tau _0}\partial ^{\alpha }_{\beta } f^{n+1}+ \sum _{i=1}^3 \partial ^{\alpha +\bar{k}_i}_{\beta -k_i}\partial ^{\alpha }_{\beta } f^{n+1}&= \frac{1}{\tau _0}\partial _{\beta }P_{Pr}(\partial ^{\alpha }f^n)+\partial ^{\alpha }_{\beta }\Gamma (f^n),\) 

and taking product with \( \partial ^{\alpha }_{\beta } f^{n+1}\)  yields
\(\frac{1}{2}\frac{d}{dt}\Vert \partial ^{\alpha }_{\beta } f^{n+1}\Vert _{L^2_{x,v}}^2+\frac{1}{\tau _0}\Vert \partial ^{\alpha }_{\beta } f^{n+1}\Vert _{L^2_{x,v}}^2 \le \sum _{i=1}^3 \int _{\mathbb {R}^3\times \mathbb {T}^3}\partial ^{\alpha }_{\beta } f^{n+1}\partial ^{\alpha +\bar{k}_i}_{\beta -k_i}\partial ^{\alpha }_{\beta } f^{n+1}dvdx \cr + \frac{1}{\tau _0}\int _{\mathbb {R}^3\times \mathbb {T}^3}(\partial ^{\alpha }_{\beta } f^{n+1}\partial _{\beta }P_{Pr}(\partial ^{\alpha }f^n)+\partial ^{\alpha }_{\beta } f^{n+1}\partial ^{\alpha }_{\beta }\Gamma (f^n))dvdx,\) 
where \(k_i~(i=1,2,3)\)  are coordinate unit vectors and \(\bar{k}_1=(0,1,0,0)\) , \(\bar{k}_2=(0,0,1,0)\) , \(\bar{k}_3=(0,0,0,1)\) .
We then integrate for time and recall Proposition REF  to derive
\((1-CT_*)\mathcal {E}(f^{n+1})(t) \le \left( \frac{1}{2}+CT_*+CT_*M_0^2\right)M_0.\) 

For sufficiently small \(M_0\)  and \(T_*\) , we conclude that
\(\mathcal {E}(f^{n+1})(t) \le M_0.\) 

The remaining part can be obtained from the standard argument [1]}, [2]}, [3]}. We omit it.
